Problem: For real numbers $x > 1,$ find the minimum value of
\[\frac{x + 8}{\sqrt{x - 1}}.\]
Answer: Let $y = \sqrt{x - 1}.$  Then $y^2 = x - 1,$ so $x = y^2 + 1.$  Then
\[\frac{x + 8}{\sqrt{x - 1}} = \frac{y^2 + 9}{y} = y + \frac{9}{y}.\]By AM-GM,
\[y + \frac{9}{y} \ge 6.\]Equality occurs when $y = 3,$ or $x = 10,$ so the minimum value is $\boxed{6}.$